Sandia is a multiprogram laboratory operated by Sandia Corporation, a Lockheed Martin Company,for the United States Department of Energy under contract DE-AC04-94AL85000.

Reconnect ‘04LP-Based Approximation Algorithms

Cynthia PhillipsSandia National Laboratories

Slide 2

Linear Programming (LP) Relaxation-Based Approximation

• Variables can take rational values (relax integrality constraints)• Efficiently solvable: gives lower bound on optimal IP solution• Common technique:

– Use structural information from LP solution to find feasible IP solution

– Bound quality using LP bound• Integrality gap = (best IP solution)/(best LP solution)• This technique cannot prove anything better than integrality gap

Slide 3

Integer Program (IP) for capacitated network design

A simple IP for capacitated network design:

Where d(C) is the maximum demand di for any pair that crosses cut C

xe = 1 if edge e is selected

min cexeΣe ∈C

uexe ≥d(C) ∀ cutsetCΣe ∈C

xe ∈ 0,1

Slide 4

Knapsack Cover (KC) Inequalities

€

u(A) = uA < D(C)e∈A∑

residual D(A) = D − u(A)uA (e) = min(ue,D(A))

KC : uA(e)xe ≥ D(A)e∈C−A∑

AC

Slide 5

Finding An Approximate Solution

Let

Set of edges at least half selected by LP• Select all these edges

– Increases cost (for A) by factor of 2• Now much meet demand D(A) = D - u(A) with rest of edges

€

A = e ∈ E | xe ≥ 12

⎧ ⎨ ⎩

⎫ ⎬ ⎭

Slide 6

Finding an Approximate Solutions

• Sort edge by ue

Consider the three cases€

uA (e) = min(ue ,D(A))if ue1

≤ ue2 then

uA (e1) ≤ uA (e2)

€

ue1≤ ue2

≤ DA

ue1≤ DA ≤ ue2

DA ≤ ue1≤ ue2

Slide 7

Finding an Approximate Solution

xe = q/p rational

r is least common multiple of denominators so rxe integral for all e

Make 2rxe “copies” of xe

(convex multipliers will be 1/r)

Slide 8

Approximate solution for knapsack (gap 2)

• 2rxe copies of edge e, sorted by capacity

• Place in r buckets, round robin

• Each bucket will be a solution Si

• No edge in any solution twice

e1 e1 e1 e1 e1 e1

e2e2

e2e2

e3e3

€

xe < 12

2rxe < r

Slide 9

All buckets are Feasible

First Bucket (biggest) Last Bucket (smallest)

ek4

ek2

e1

ek3

ek1

<

<

€

uA (e)e∈first∑ ≤ uA (e)

e∈last∑ + D(A)

Slide 10

All Buckets Feasible

Suppose

We have

So for all buckets

From total capacity:

Contradicts KC inequality

€

uA (e)e∈last∑ < D(A)

€

uA (e)e∈first∑ ≤ uA (e)

e∈last∑ + D(A)

€

uA (e)∑ < 2D(A)

€

uA (e)e∈E−A∑ xe < 2rD(A)

Slide 11

Separation

Only have to satisfy KC inequality for

Add these cuts if violated till we get an LP solution where KC inequality holds for it’s A.

€

A = e ∈ E | xe ≥ 12

⎧ ⎨ ⎩

⎫ ⎬ ⎭

Slide 12

Polynomial Time

Really only m+1 distinct solutions

e1 e1 e1 e1 e1 e1

e2e2

e2e2

e3e3

Slide 13

A Scheduling Example

Given n jobs J1, J2, …, Jn

Job Ji has length pi, weight wi

Precedence constraints: mean Ji must finish before Jj starts

No preemption, one machine

Cj = completion time of job Jj

Goal: minimize

NP-complete. We’ll get a 4-approximation

J i p J j

wjC jj=1

n∑

Slide 14

Integer Programming Formulation

Subject to

xjt = 1 if job Jj completes at time t0 otherwise

⎧ ⎨ ⎩

xjtt=p j

T∑ =1 ∀j

xju ≤1u=t

t+pj −1

∑j=1

n∑ t =1,...,T = pjj=1

n∑xju

u=1

t∑ − xkuu=1

t+pk∑ ≥0 ∀J i p J k, t =1,...,T −pk

xjt ∈ 0,1{ } j =1,...,n; t =pj ,...,T

€

min w jt= p j

T

∑j=1

n

∑ tx jt

Slide 15

Constraint: One Job at a Time

Consider all (job, finish time) pairs that would run over (t-1, t]

xju ≤1u=t

t+pj −1

∑j=1

n∑t

t+1

t+2

t+pj-1

t-1

T-pj

...

t-1

Slide 16

Precedence Constraints

If job Jk finishes by time t + pk, then job Jj must finish by time t

xjuu=1

t∑ − xkuu=1

t+pk∑ ≥0 J j p J k

Slide 17

LP relaxation, Fractional Schedule

xjt

pj

Slide 18

Fractional Schedule x*

Fractional Completion Time:

Midpoint: min t* such that

C j* = txjt

*

t=pj

T∑

xjt*

t=p j

t*

∑ ≥12

Slide 19

Approximation Algorithm

• Solve LP• Compute midpoints for all jobs• Order by midpoints

Slide 20

Approximate Schedule is feasible

• No preemption• One job at a time• Precedence constraints

Midpoint of Jj < Midpoint of Jk

xjuu=1

t∑ − xkuu=1

t+pk∑ ≥0 J j p J k

Slide 21

Proof of Quality Road Map

• Relate Cj to LP values

Renumber jobs by midpoint:

We’ll show

t1* ≤t2

* ≤...≤tn*

C j ≤2tj* and

C j* ≥tj

*

2 ⇒ Cj ≤4C j

* ⇒ 4-approximation

Slide 22

Upper Bound on Completion Times

• At time tj* fractional schedule has done pj/2 work.

• Since tk* tj* for k<j, schedule has done pk/2 work on Jk.

• One unit of work/time unit • But by construction

t-pjt

xjt

total≥12 pk

k=1

j

∑tj* ≥1

2 pkk=1

j

∑C j = pk

k=1

j

∑⇒ Cj ≤2tj

*

Slide 23

Lower Bound on LP values

• By definition:

So C j* = txjt

t=pj

T∑ ≥ txjt

t=tj*

T

∑ ≥tj

* xjtt=tj

*

T∑≥1

2tj*

xju <12u=pj

tj* −1

∑

Slide 24

Proof of Quality

Therefore

C j ≤2tj* and

C j* ≥tj

*

2 ∀j ⇒ Cj ≤4C j

* ∀j

wjCj ≤4 wjC j*∑

j∑ ≤4 wjtxjt

*

j∑

≤4* Optimal

Slide 25

Comments

• Can create alternative schedules using point tj

• LP-based approximation algorithms can give feasible solutions in branch and bound

• Other LP-based approximation algorithms for scheduling problems are based on matching/assignment

tjα = min t s.t. xjt'

t'=pj

t∑ ≥α